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DNS computational requirements

It was mentioned in the theory section that the computational cost of DNS was very high. This approach does not rely on any assumption or turbulence model. The whole range of temporal and spatial scales must be resolved. This means that the mesh must be large enough to capture the largest eddies, with a resolution sufficient to resolve the smallest scales. From dimensional analysis, it is possible to estimate the size of these smallest scales (Kolmogorov length scale η), at which the turbulent kinetic energy is dissipated by viscosity.
(1) η = (ν3)/(ϵ)1 ⁄ 4
The energy is dissipated at the small scales, but the rate at which this energy is dissipated is ultimately governed by the large eddies (integral length scale l). If u is some characteristic turbulent velocity at the large scales, the dissipation can therefore be obtained from
(2) ϵ ~ (u3)/(l)
This equation is supported by a large set of experimental data and numerical simulations [1]. Combining Eq.(1↑) with Eq.(2↑), it directly follows that
(3) (l)/(η) ~ (ul)/(ν)3 ⁄ 4 =  Re3 ⁄ 4
where Re is the turbulent Reynolds number. In order to capture all length scales as described in the previous paragraph, the grid must contained at least N’ = l ⁄ η points in each dimension. For a three dimensional simulation, the total number of points is therefore
(4) N ~ Re9 ⁄ 4
Hence, the computational cost of DNS grows very fast with increasing Reynolds number. Furthermore, the time integration of the solution must be done explicitly. As a consequence, the simulation time step Δt must be small enough to ensure that the fluid particles will move only a fraction of the mesh spacing h at each step. This is typically called the Courant-Friedrichs-Lewy (CFL) condition, and can be written
(5) CFL = (uΔt)/(h) < 1
Because h is so small (comparable to the Kolmogorov length scale η), the time step Δt required to satisfy the CFL condition is very small too.
(6) Δt ~ (η)/(u)
Typically, the total simulation time is roughly proportional to the turbulence length scale τ ~ L ⁄ u. It follows that the total number of time steps required for the calculation is
(7) Nt = (τ)/(Δt) ~ (l)/(η)
Eventually, it appears that the total number of operations in the simulation grows as Re9 ⁄ 4 Re3 ⁄ 4 =  Re3. This simple relation explains why DNS is so computationally expensive, even at low Reynolds number.

LES filter

In the theory section, the general principle of LES was explained. A filter is applied to the Navier-Stokes equations in order to discard the scales that will be judged “small enough" to be modeled. For example, the filtered velocity is
(8) ui(x, t) = GΔ(x − ζ)ui(ζ, t)d3 ζ
where Δ is the filter width and G is a function (Kernel) whose value approaches zero if ui occurs on a length scale smaller than Δ. A typical example of convolution kernel is the Gaussian filter (in three dimensions):
(9) GΔ(x) = (6)/(πΔ2)3 ⁄ 2exp − (6∥ x2)/(Δ2)
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